User talk:Hyp cos/Linear Array Notation
Extension Make an extension. {\(\Omega^{n+1}\)} = {\(\Omega^{n}\Omega\)} {\(\Omega\)} = {0,1} {\(\Omega 2\)} = {0,2} {\(\Omega^2\)} = {0,0,1} {\(\Omega^3\)} = {0,0,0,1} Also \(\Omega^\omega\) = \(\Omega\){0}. ...AarexTiao 12:49, November 15, 2013 (UTC) :Of course, without extensions googology can't exist. Ikosarakt1 (talk ^ ) 17:05, November 15, 2013 (UTC) Niceties Complicated stuff... I'm interested how some inconvenient ordinals like \(\theta(\omega^{\Omega+\theta(2)})\) must be expressed in this notation. Ikosarakt1 (talk ^ ) 17:03, November 15, 2013 (UTC) Strength It is \(\psi(\Psi_{\Xi(\omega,0)}(0))\). The second entry works like \(\Omega\), the third like \(I\), the fourth like \(M\), the fith like \(\Xi(3,0)\), etc. Even I am not very sure about its strength. It's a little hard to compare with FGH (or other notations), because Subrule 2 works so strange - From {0,{0}} up, we get things like 1{0,{0}}, {0}{0,{0}}, {0,{0}}, {0,{0}}, {0,{0}},...easily. But things suddenly go wrong. nR{0,1}{0,{0}} expand to nR }{0,{0}}}...{0,{0}}}{0,{0}}}=nR {0,{0}}}...{0,{0}}}{0,{0}}}=nR {0,{0}}}...{0,{0}}}{0,{0}}},n-1}=... in this "expanding process" we don't use Rule 2, so the base number n doesn't change. We add an encloser { and ,n-1} then expand it, add an encloser { and ,n-2} then expand it, ... add an encloser { and ,1} then expand it! And when n changes by Rule 2 we must do this again. This will force the notation to go a big step. There's a big gap between {0,{0}} and ,gongulus}...,2},1}} in growth rate, but not in numbers. Every expression in R function can be worked out since my R function is well-defined. It can be worked out in numbers, but it's "uncomparable" to other notations - How strange! hyp$hyp?cos&cos (talk) 01:16, November 16, 2013 (UTC) :{0,0,{0}} has level \(\psi(\psi_{I_\omega}(0))\) :{0,0,{0,0,1}} has level \(\psi(\psi_{I_I}(0))\) :{0,0,0,1} has level \(\psi(\psi_{I(1,0)}(0))\) :{0,0,n,1} has level \(\psi(\psi_{I(1,n)}(0))\) :{0,0,{0,0,1},1} has level \(\psi(\psi_{I(1,I)}(0))\) :{0,0,{0,0,{0,0,0,1}},1} has level \(\psi(\psi_{I(1,I(1,0))}(0))\) :{0,0,{0,0,0,1},1} has level \(\psi(\psi_{I(2,0)}(0))\) AarexTiao 22:10, December 11, 2013 (UTC) {0,0,{0,1,0,1},1} has level \(\psi(\psi_{I(M,0)}(0))\) AarexTiao 00:39, December 19, 2013 (UTC) Ordinal for {0, 0, 1} I'm going to ask the same question I asked about Wythagoras's Extended Bracket Notation: is it clear that {0, 0, 1} reaches \(\psi_\Omega(\psi_I(0))\), and not \(\psi_\Omega(\Omega_\Omega)\), which is the first ordinal \(\alpha\) such that \(\alpha = \psi_\Omega(\Omega_\alpha)\)? Wythogoras's notation manages to surpass that ordinal, because the levels of levels can seek out higher level brackets, so for example in [0_{0_2}], the 0_2 seeks out the outermost bracket as the next level down, and it evaluates to [0_{[0_{[0_...]}]}]. But if we take say {0, {0, 2}} in Hyp Cos's notation, I believe we would evaluate that to {0, , and that would then nest in the first argument rather than the second. So that suggests that {0, {0, 2}} would correspond to \(\psi_\Omega(\Omega_{\text{BHO}})\), and in general if X corresponds to the ordinal \(\alpha\) then {0, X} corresponds to \(\psi_\Omega(\Omega_\alpha)\). So then {0, 0, 1} would correspond to \(\psi_\Omega(\Omega_\Omega)\). Deedlit11 (talk) 17:32, June 28, 2014 (UTC) {0, } is \(\psi(\Omega_{\varepsilon_0})\), {0, } is \(\psi(\Omega_{BHO})\), and {0,{0,1}} is \(\psi(\Omega_\Omega)\), {0,{0,2}} is \(\psi(\Omega_{\Omega_2})\). 3R{0, }=3R{0, }=3R{0, }=... 3R{0,{0,1}}=3R }=3R }}}}=3R }}}=... 3R{0,{0,2}}=3R ,1}=3R ,1}},1}=3R ,1}},1}=... Note that {0, ____ } has higher level than { ____ ,1} and { ____ }. hyp$hyp?cos&cos (talk) 01:18, June 29, 2014 (UTC) :Okay, thanks. I still have some trouble understanding Rule 5 and Subrule 2 though: they seem to assume that you will find an encloser less than or equal to {⊘1 ____ ,a⊙1}, but what if there is no such encloser? Your above examples seem to indicate that you place an encloser on the outside, but you should clarify that in the rules. :Also, to clarify your rules on comparing levels of braces: does {1{0}1} have the same level as {1{0}2}? :Finally, do you plan on extending your R notation? Deedlit11 (talk) 02:16, June 29, 2014 (UTC) :Oh, there's a problem if we cannot find an encloser. Now it's fixed. :And yes, {1{0}1} have the same level as {1{0}2}. Generally, in level comparisons, a high-level brace (not a separator) can "eat" lower-level braces after it. e.g. {3{1}{0}2{0}1{1}2{0}{0}} has the same level as {3{1}{1}}. :Finally, I think it's not easy to extend the notation. My HNAN seems a bit like Bird's array notation - :*Both use "nested levels", in BAN the /k must be inside at least k layer of brackets, and in HNAN the ,,...(k)...,, must be inside at least k layer of braces. :*It needs some work to see the "nested levels". :*The rules on the ,,...(k)...,, are related to not only (k-1)-separators, but also (k-2)-separators, ... , 2-separators and 1-separators. The rules on the /k are related to not only (k-1)-hyperseparators, but also (k-2)-hyperseparators, ... , 2-hyperseparators and 1-hyperseparators. That's the point. :So I think the extension will be very complex but not strong (See "Beyond nested array notation V" for example). And I want to start a new version of it. hyp$hyp?cos&cos (talk) 05:36, June 29, 2014 (UTC) Ordinal for {0,1}{0,1} If {0,1}{0,1} is the ordinal for \varepsilon_1 , then what's the expression for \varepsilon_0*2 ? Also, picking the rightmost {0,1} as \odot_2 and resolving leftmost {0,1} gives something like {}{0,1} (n {}'s). Ikosarakt1 (talk ^ ) 16:24, July 3, 2014 (UTC) :Perhaps ? Deedlit11 (talk) 16:44, July 3, 2014 (UTC) :In both examples, the right {0,1} (or ) can be taken as \odot_2 and the left resolves, by rule 5, to }...}}} (with n+1 {}'s). So they both equal to \varepsilon_0*2 . By the similar reasoning, {0,1}{0,1}...{0,1}{0,1} (m {0,1}'s) = \varepsilon_0*m and {1,1} is just \omega^{\varepsilon_0+1} . There are some conjectures: {a,1} = \omega^{\epsilon_0+a} {0,2} = \varepsilon_1 {0,1+a} = \varepsilon_\alpha {0,0,1} = \zeta_0 {0,0,0,1} = \eta_0 {0 {1*} 1} = \phi(\omega,0) Am I wrong? Ikosarakt1 (talk ^ ) 17:05, July 3, 2014 (UTC) :The point is that when we resolve {0, 1}, we have to find the nearest encloser that encloses {0, 1} and has a level less than or equal to {_, 0}. There isn't one, so we add such an encloser enclosing the entire notation. So in {0, 1} {0, 1}, the second {0, 1} becomes \odot_3 , not \odot_2 , and the expression resolves as {0,1}}...{0,1}}. Deedlit11 (talk) 17:36, July 3, 2014 (UTC) \(\psi(\psi_{\Xi(\omega,0)}(0))\) I always thought that the limit of the sequence \chi(1,0), \chi(2,0), \chi(3,0) is just \chi(\omega,0) and not \psi_{\Xi(\omega,0)}(0) . If you use your own ruleset, please write it down. Ikosarakt1 (talk ^ ) 17:12, July 3, 2014 (UTC)